Sara is an avid lottery player. In the certain game...

Hello LUANDATO.

Let's take a look at your question.

We have to find all the prime factors of each option.

Option A: $$32=2^5\ \ \ -\ \ \ 48=2^4\cdot3\ \ \ -\ \ \ 52=2^2\cdot13.$$ Option B: $$33=3\cdot11\ \ \ -\ \ \ 42=2\cdot3\cdot7\ \ \ -\ \ \ 56=2^3\cdot7.$$ Option C: $$39=3\cdot13\ \ \ -\ \ \ 40=2^3\cdot5\ \ \ -\ \ \ 54=2\cdot3^3.$$ Option D: $$38=2\cdot19\ \ \ -\ \ \ 49=7^2\ \ \ -\ \ \ 51=3\cdot17.$$ Option E: $$36=2^2\cdot3^2\ \ \ -\ \ \ 42=2\cdot3\cdot7\ \ \ -\ \ \ 56=2^3\cdot7.$$ If we take a look of all distinct prime factors on each set, we can conclude that the option D has 5 distinct prime factors and the rest has 4 distinct prime factors or less.

Hence, the correct answer is the option D. I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Regards.